[{"id":998,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生调查了班级同学最喜欢的运动项目，收集数据后制作了频数分布表。其中喜欢跳绳的有8人，喜欢踢毽子的有5人，喜欢跑步的有12人，喜欢打篮球的有15人。则喜欢打篮球的人数占总人数的百分比是______%。","answer":"37.5","explanation":"首先计算总人数：8 + 5 + 12 + 15 = 40（人）。喜欢打篮球的人数为15人，因此所占百分比为 (15 ÷ 40) × 100% = 37.5%。本题考查数据的收集、整理与描述中的百分比计算，属于简单应用。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:50:48","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":793,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生测量了教室里5个不同位置的气温，分别为-2℃、3℃、0℃、-5℃和4℃，这些气温的平均值是___℃。","answer":"待完善","explanation":"首先将所有气温相加：-2 + 3 + 0 + (-5) + 4 = 0。然后将总和除以数据的个数5，得到平均值为0 ÷ 5 = 0。因此，这些气温的平均值是0℃。本题考查有理数的加减运算及平均数的计算方法，属于数据的收集、整理与描述知识点，符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:09:30","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1476,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加环保知识竞赛，竞赛成绩以百分制记录。为分析成绩分布情况，某学生随机抽取了50名参赛学生的成绩，整理后得到如下信息：成绩在60分以下的有5人，60～69分的有8人，70～79分的有12人，80～89分的有15人，90～100分的有10人。已知所有被抽取学生的平均成绩为78.6分，且90～100分这一组中，最低分为92分，最高分为100分，该组平均分为96分。若将80～89分这一组的所有成绩都提高5分，同时将60～69分这一组的所有成绩都降低3分，其余组数据不变，求调整后这50名学生的平均成绩（精确到0.1分）。","answer":"解题步骤如下：\n\n第一步：计算原始总分。\n已知平均成绩为78.6分，总人数为50人，\n所以原始总分 = 78.6 × 50 = 3930（分）。\n\n第二步：计算90～100分组原始总分。\n该组有10人，平均分为96分，\n所以该组原始总分 = 96 × 10 = 960（分）。\n\n第三步：计算其余四组的原始总分。\n其余四组总人数 = 50 - 10 = 40人，\n其余四组原始总分 = 3930 - 960 = 2970（分）。\n\n第四步：分析调整情况。\n- 60～69分组：8人，每人成绩降低3分，总分减少 8 × 3 = 24（分）。\n- 80～89分组：15人，每人成绩提高5分，总分增加 15 × 5 = 75（分）。\n- 其他组（60分以下、70～79分、90～100分）成绩不变，总分不变。\n\n第五步：计算调整后总分。\n调整后总分 = 原始总分 - 24 + 75 = 3930 + 51 = 3981（分）。\n\n第六步：计算调整后平均成绩。\n调整后平均成绩 = 3981 ÷ 50 = 79.62（分）。\n精确到0.1分，结果为79.6分。\n\n答：调整后这50名学生的平均成绩为79.6分。","explanation":"本题综合考查了数据的收集、整理与描述中的频数分布、平均数计算，以及有理数的混合运算和一元一次方程思想的应用（虽未显式列方程，但总分与平均数的关系本质上是线性关系）。解题关键在于理解平均数与总分之间的转换，并能准确计算各组调整对总分的影响。题目设置了真实情境，要求学生在多组数据中识别变化部分，排除干扰信息（如90～100分组的详细数据仅用于验证，实际解题中只需其总分），体现了数据分析能力和逻辑推理能力。难度较高，因涉及多步运算、信息筛选和精确计算，符合困难级别要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:53:43","updated_at":"2026-01-06 11:53:43","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":599,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间时，收集了10名同学每周阅读课外书的时间（单位：小时），数据如下：3, 5, 4, 6, 3, 7, 5, 4, 5, 6。为了分析数据，该学生计算了这组数据的平均数，并发现如果将每个数据都增加2小时，新的平均数比原来多2小时。现在，该学生想进一步了解数据分布情况，于是他绘制了一个条形统计图。以下关于这组数据的说法中，正确的是：","answer":"A","explanation":"首先将原始数据从小到大排列：3, 3, 4, 4, 5, 5, 5, 6, 6, 7。共有10个数据，为偶数个，因此中位数是第5个和第6个数据的平均数，即(5 + 5) ÷ 2 = 5，所以A正确。众数是出现次数最多的数，5出现了3次，是最多的，因此众数是5，B错误。平均数计算为：(3+5+4+6+3+7+5+4+5+6) ÷ 10 = 48 ÷ 10 = 4.8，C错误。极差是最大值减最小值：7 - 3 = 4，D错误。因此正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:01:12","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"这组数据的中位数是5小时","is_correct":1},{"id":"B","content":"这组数据的众数是6小时","is_correct":0},{"id":"C","content":"这组数据的平均数是4.5小时","is_correct":0},{"id":"D","content":"这组数据的极差是3小时","is_correct":0}]},{"id":678,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在平面直角坐标系中，点 A 的坐标是 (3, -2)，点 B 位于点 A 的正上方 5 个单位长度处，则点 B 的坐标是 ___","answer":"(3, 3)","explanation":"点 A 的坐标是 (3, -2)，表示横坐标为 3，纵坐标为 -2。点 B 在点 A 的正上方 5 个单位长度，说明横坐标不变，纵坐标增加 5。因此，点 B 的纵坐标为 -2 + 5 = 3，横坐标仍为 3，所以点 B 的坐标是 (3, 3)。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:26:54","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1939,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生调查了班级同学每周用于体育锻炼的时间（单位：小时），将数据整理后发现，锻炼时间在4小时及以下的有12人，5小时的有8人，6小时的有x人，7小时的有y人。已知这组数据的平均数为5.5小时，且众数为6小时，则x + y的值为____。","answer":"15","explanation":"由众数为6知x最大；设总人数为30+x+y，列平均数方程：(12×4+8×5+6x+7y)\/(30+x+y)=5.5，化简得x+1.5y=15。因x>8且为整数，试值得x=9,y=4不满足，x=6,y=6不满足，x=3,y=8时x非最大，最终x=12,y=2满足条件，x+y=14？重新计算：正确解为x=12,y=2不满足众数，实际x=9,y=4时x=9>8成立，x+y=13？更正：正确解为x...","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:11:19","updated_at":"2026-01-07 14:11:19","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":338,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"150","answer":"答案待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:40:21","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":460,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"144度","answer":"待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:48:43","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":2313,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园绿化项目中，工人师傅需要在一块矩形空地的对角线上铺设一条石板路。已知这块空地的长为8米，宽为6米。为了估算石板数量，需要先计算对角线的长度。根据勾股定理，这条对角线的长度最接近以下哪个值？","answer":"B","explanation":"本题考查勾股定理在矩形对角线计算中的应用。矩形对角线将矩形分成两个直角三角形，其中两条直角边分别为矩形的长和宽。根据勾股定理：对角线² = 长² + 宽² = 8² + 6² = 64 + 36 = 100。因此，对角线 = √100 = 10（米）。故正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:46:09","updated_at":"2026-01-10 10:46:09","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"9米","is_correct":0},{"id":"B","content":"10米","is_correct":1},{"id":"C","content":"11米","is_correct":0},{"id":"D","content":"12米","is_correct":0}]},{"id":1444,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加数学实践活动，要求每名学生从A、B、C三个任务中至少选择一个完成。已知共有120名学生参与，其中选择A任务的有78人，选择B任务的有65人，选择C任务的有52人。同时，恰好选择两个任务的学生人数是恰好选择三个任务学生人数的3倍，且没有学生一个任务都不选。问：恰好选择三个任务的学生有多少人？","answer":"设恰好选择三个任务的学生人数为x人。\n\n根据题意，恰好选择两个任务的学生人数是3x人。\n\n因为每个学生至少选择一个任务，所以所有学生可以分为三类：\n- 只选一个任务的：设为y人\n- 恰好选两个任务的：3x人\n- 恰好选三个任务的：x人\n\n总人数为120人，因此有：\ny + 3x + x = 120\n即：y + 4x = 120 ——（1）\n\n再从任务被选的总人次角度分析：\n- 选择A任务的有78人，B任务65人，C任务52人，总人次为：78 + 65 + 52 = 195\n\n每个只选一个任务的学生贡献1人次，\n每个选两个任务的学生贡献2人次，\n每个选三个任务的学生贡献3人次。\n\n因此总人次可表示为：\n1×y + 2×(3x) + 3×x = y + 6x + 3x = y + 9x\n\n所以有：y + 9x = 195 ——（2）\n\n用方程（2）减去方程（1）：\n(y + 9x) - (y + 4x) = 195 - 120\n5x = 75\n解得：x = 15\n\n代入（1）得：y + 4×15 = 120 → y = 60\n\n因此，恰好选择三个任务的学生有15人。\n\n答：恰好选择三个任务的学生有15人。","explanation":"本题考查数据的收集、整理与描述中的集合思想与方程建模能力，结合一元一次方程和二元一次方程组的解法。解题关键在于理解“人次”与“人数”的区别，并合理设未知数，建立两个不同角度的等量关系：一是总人数，二是任务被选的总人次。通过设恰好选三个任务的人数为x，利用“恰好选两个任务的人数是其3倍”建立联系，再结合总人数和总人次列出方程组，最终求解。本题综合性强，需要学生具备较强的逻辑分析和方程建模能力，属于困难难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:41:23","updated_at":"2026-01-06 11:41:23","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]}]